Question about this Continuity Equation (electromagnetism)

[hagopbul]

Hello All :

reading the Bo Thide book in electromagnetism , downloaded the draft copy from the following link http://www.plasma.uu.se/ , i reached the chapter 4 now and a section in that chapter (section 4.3) have few lines that i coudnt understand (mathematically speaking)

the writer conclude an continuity equation that have two parts the left part is the change in particlees density into the volume d(ρ)/dt
and divergence of current density of charges ∇. j = ∇.(ρv) : ρ charge density ,v velocity of charges , j current density of charges

the right side of the equation there is s which is the density of charges generated in the volume V that we are studying

my question how the left side and the right side are equal to each other after all the equation assume that the current coming into the volume is equal to the charges density generated in that volume

what if the sides of the equation is not equal what should we add to the right side of the equation or the left side

the question from 2nd year physics (maybe) i am currently reviewing and couldn't remember this chapter correctly

best
H.B.

http://www.plasma.uu.se/CED/Book where i downloaded the book from

 

[vanhees71]

For some reason, the link to the book doesn't work.

The continuity equation reads
$$\partial_t \rho =-\vec{\nabla} \cdot \vec{j}=-\vec{\nabla} (\rho \vec{v}),$$
where $\rho$ is the charge density and $\vec{j}=\rho \vec{v}$ the current density and $\vec{v}$ is the flow field of the charges.

To understand it, it's easier to consider the integral form. To that end we consider an arbitrary volume $V$, at rest (in the inertial reference frame we are working in). Then the charge contained in this volume is
$$Q_V(t)=\int_{V} \mathrm{d}^3 x \rho(t,\vec{x}).$$
Now the continuity equation expresses nothing else than the conservation of electric charge. This means that the charge inside the volume can only change with time by charge flowing out of or into the volume.

To calculate the amount of charge flowing through the boundary surface $\partial V$ of the medium we define the surface normal vectors $\mathrm{d}^2 \vec{f}$ pointing out of the medium at any point on $\partial V$. Now in an infinitesimal time interval $\mathrm{d} t$ at time $t$ the charge contained in a volume element $\mathrm{d} V=\mathrm{d} t \vec{v} \cdot \mathrm{d}^2 \vec{f}$
flows out of the medium (counting positive when the flow is really outwards, i.e., when the angle between $\mathrm{d}^2 \vec{f}$ and $\vec{v}$ is in the interval $[0,\pi/2]$ and is in fact inwards when this angle is in the interval $[0,\pi/2]$.

Now the change of the charge in the volume obviously is
$$\mathrm{d} t \dot{Q}(t)=\int_{V} \mathrm{d}^3 x \partial_t \rho(t,\vec{x})=-\mathrm{d} t\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{v}(t,\vec{x} \rho(t,\vec{x}).$$
The $-$ sign on the right-hand side is, because charge flowing out gets out of (flows into) the volume has to be counted as negative (positive), i.e., always opposite to the sign of $\mathrm{d} V$.

Now you can use Gauß's integral law to write
$$\dot{Q}(t)=\int_{V} \mathrm{d}^3 x \partial_t \rho(t,\vec{x})=-\int_V \mathrm{d}^3 x \vec{\nabla} \cdot [\rho(t,\vec{x})\vec{v}(t,\vec{x})].$$
Further since this holds for any volume, we can make the volume as small we like around the point $\vec{x}$, and then from the definition of the divergence operator we find the local form of charge conservation,
$$\partial_t \rho=-\vec{\nabla} (\rho \vec{v}).$$
Further it's clear that $\vec{j}=\rho \vec{v}$ describes the charge per unit time flowing through a surface element with normal vector $\mathrm{d}^2 \vec{f}$ by $\vec{d}^2 \vec{f} \cdot \vec{v}$, and thus $\vec{j}$ is the electric-charge density.

 

[hagopbul]

thanks man but what i am talking about is slightly different, look into page 54
another link for the book (draft)
http://scipp.ucsc.edu/~haber/ph214/EMFT_Book_Thide.pdf

 

[vanhees71]

For electric charge $s$ is exactly 0 from very fundamental principles (gauge invariance). If it turns out not to be that would make all standard electrodynamics obsolete!